Factor the quadratic expression completely. $2x^2+7x+3=$
Explanation: Since the terms in the expression do not share a common monomial factor and the coefficient on the leading $x^2$ term is not $1$, let's factor by grouping. The expression ${2}x^2{+7}x{+3}$ is in the form ${A}x^2+{B}x+{C}$. First, we need to find two integers ${a}$ and ${b}$ such that: $\begin{cases} &{a}+{b}={B}={7} \\\\ &{ab}={A}{C}= ({2})({3})=6 \end{cases}$ We find that ${a}={1}$ and ${b}={6}$ satisfy these conditions, since ${1}+{6}={7}$ and $({1})({6})=6$. Next, we can use these values to rewrite the $x$ -term and factor by grouping. $\begin{aligned} 2x^2+7x+3&=2x^2+{6}x+{1}x+3 \\\\ &=2x(x+3)+1(x+3) \\\\ &=(x+3)(2x+1) \end{aligned}$ In conclusion, $2x^2+7x+3=\left(x+3\right)\left(2x+1\right)$